then we can assume that when the equation has infinite terms the answer is 2.
and reaches the following "proof" and let me know if it's right:
1 image:https://gyazo.com/49a46e56fb19b4ec7aa21594e4e78cd1 [Broken]
2 image:https://gyazo.com/c6b99485e6d0271c1c0bbdbaaca29d54 [Broken]

Besides knowing if this is OK too I wonder if this is what is called "inductive method"

I don't like your proof! If you have
x^n = 1 + x+ x^2 +...+x^(n-1)
using the formula for the sum of a GP you get
x^n = (1-x^n)/(1-x), and so
x - 2 + 1/x^n =0
now you can see that as n tends to infinity x gets as close as you like to 2.

I don't like your proof! If you have
x^n = 1 + x+ x^2 +...+x^(n-1)
using the formula for the sum of a GP you get
x^n = (1-x^n)/(1-x), and so
x - 2 + 1/x^n =0
now you can see that as n tends to infinity x gets as close as you like to 2.

I have a question.
How did you get in here
x^n = (1-x^n)/(1-x)
to here
x - 2 + 1/x^n =0
my poor mind can not compute